The function $f ( x , y ) = x ^ { 2 } + y ^ { 2 } + 3 x y$ has one critical point. Determine its location and type.
A) $( 2,1 )$ , saddle point
B) $( 2,1 )$ , minimum point
C) $( 2,1 )$ , maximum point
D) $( \sqrt { 2 } , 1 )$ , saddle point
E) $( \sqrt { 2 } , 1 )$ , minimum point
F) $( \sqrt { 2 } , 1 )$ , maximum point
G) $( 0,0 )$ , saddle point
H) $( 0,0 )$ , minimum point

Question

Quizplus · Accepted Answer

To find the critical points of $f(x, y) = x^2 + y^2 + 3xy$, we first find the partial derivatives and set them equal to zero:$$f_x = 2x + 3y = 0$$$$f_y = 2y + 3x = 0$$Solving this system of equations, we find $x = 0$ and $y = 0$, so the critical point is at $(0,0)$.To determine the type of critical point, we use the second derivative test, which involves the Hessian matrix $H$:$$H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 3 \ 3 & 2 \end{bmatrix}$$The determinant of $H$ at $(0,0)$ is $2*2 - 3*3 = 4 - 9 = -5$, which is less than 0, indicating a saddle point.

The Function f(x,y)=x2+y2+3xyf ( x , y ) = x ^ { 2 } + y ^ { 2 } + 3 x yf(x,y)=x2+y2+3xy

The Function $f ( x , y ) = x ^ { 2 } + y ^ { 2 } + 3 x y$